Question: Is ${483782}$ divisible by $3$ ?
A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {483782}= &&{4}\cdot100000+ \\&&{8}\cdot10000+ \\&&{3}\cdot1000+ \\&&{7}\cdot100+ \\&&{8}\cdot10+ \\&&{2}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {483782}= &&{4}(99999+1)+ \\&&{8}(9999+1)+ \\&&{3}(999+1)+ \\&&{7}(99+1)+ \\&&{8}(9+1)+ \\&&{2} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {483782}= &&\gray{4\cdot99999}+ \\&&\gray{8\cdot9999}+ \\&&\gray{3\cdot999}+ \\&&\gray{7\cdot99}+ \\&&\gray{8\cdot9}+ \\&& {4}+{8}+{3}+{7}+{8}+{2} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${483782}$ is divisible by $3$ if ${ 4}+{8}+{3}+{7}+{8}+{2}$ is divisible by $3$ Add the digits of ${483782}$ $ {4}+{8}+{3}+{7}+{8}+{2} = {32} $ If ${32}$ is divisible by $3$ , then ${483782}$ must also be divisible by $3$ ${32}$ is not divisible by $3$, therefore ${483782}$ must not be divisible by $3$.